/*
 * @lc app=leetcode.cn id=102 lang=cpp
 *
 * [102] 二叉树的层序遍历
 *
 * https://leetcode-cn.com/problems/binary-tree-level-order-traversal/description/
 *
 * algorithms
 * Medium (64.59%)
 * Likes:    1264
 * Dislikes: 0
 * Total Accepted:    534.5K
 * Total Submissions: 827.3K
 * Testcase Example:  '[3,9,20,null,null,15,7]'
 *
 * 给你二叉树的根节点 root ，返回其节点值的 层序遍历 。 （即逐层地，从左到右访问所有节点）。
 * 
 * 
 * 
 * 示例 1：
 * 
 * 
 * 输入：root = [3,9,20,null,null,15,7]
 * 输出：[[3],[9,20],[15,7]]
 * 
 * 
 * 示例 2：
 * 
 * 
 * 输入：root = [1]
 * 输出：[[1]]
 * 
 * 
 * 示例 3：
 * 
 * 
 * 输入：root = []
 * 输出：[]
 * 
 * 
 * 
 * 
 * 提示：
 * 
 * 
 * 树中节点数目在范围 [0, 2000] 内
 * -1000 <= Node.val <= 1000
 * 
 * 
 */

// @lc code=start
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        // queue<TreeNode*> que;
        // vector<vector<int>> res;
        // TreeNode* cur = root;
        // if(cur == NULL) return  res;

        // que.push(cur);
        // while(!que.empty()){

        //     int size = que.size();
        //     vector<int> vec;

        //     for(int i = 0; i < size; i++){
        //         cur = que.front();
        //         que.pop();
        //         vec.push_back(cur->val);
        //         if(cur->left) que.push(cur->left);
        //         if(cur->right) que.push(cur->right);
        //     }
        //     res.push_back(vec);
        // }
        // return res;


        queue<TreeNode*> que;
        vector<vector<int>> res;
        TreeNode* cur = root;

        if(cur == NULL) return res;

        que.push(cur);
        

        while(!que.empty())
        {
            int size = que.size();

            vector<int> vec;

            for(int i = 0; i < size; i++){
                cur = que.front();
                que.pop();

                vec.push_back(cur->val);
                if(cur->left) que.push(cur->left);
                if(cur->right) que.push(cur->right);
            }

            res.push_back(vec);
            
        }

        return res;

    }
};
// @lc code=end

